Cambridge IGCSE Mathematics - Additional 0606 — 2007 May/June Paper 1 · Variant 1

This document consists of 6 printed pages and 2 blank pages. SP (SLM/CGW) T35481/2 © UCLES 2007 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Write your answers on the separate Answer Booklet/Paper provided. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80. * 1 3 9 8 5 9 5 4 1 8 * ADDITIONAL MATHEMATICS 0606/01 Paper 1 May/June 2007 2 hours Additional Materials: Answer Booklet/Paper Graph paper Electronic calculator Mathematical tables www.XtremePapers.com

2 0606/01/M/J/07 © UCLES 2007 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x b b ac a = − − 2 4 2 . Binomial Theorem (a + b)n = an + ( n 1)an–1 b + ( n 2)an–2 b2 + … + ( n r)an–r br + … + bn, where n is a positive integer and ( n r) = n! (n – r)!r! . 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1. sec2 A = 1 + tan2 A. cosec2 A = 1 + cot2 A. Formulae for ∆ABC a sin A = b sin B = c sin C . a2 = b2 + c2 – 2bc cos A. ∆ = 1 2 bc sin A.

3 0606/01/M/J/07 © UCLES 2007 [Turn over 1 (a) A B C
The diagram above shows a universal set
and the three sets A, B and C. (i) Copy the above diagram and shade the region representing (A C
) B. [1] (ii) A B C
Express, in set notation, the set represented by the shaded region in the diagram above. [1] (b) X Y
The diagram shows a universal set
and the sets X and Y. Show, by means of two diagrams, that the set (X Y)
is not the same as the set X
Y
. [2] 2 Find the equation of the normal to the curve y = 2x + 4 x – 2 at the point where x = 4. [5] 3 The straight line 3x = 2y + 18 intersects the curve 2x2 – 23x + 2y + 50 = 0 at the points A and B. Given that A lies below the x-axis and that the point P lies on AB such that AP : PB = 1 : 2, find the coordinates of P. [6] 4 (i) Find the first three terms, in ascending powers of u, in the expansion of (2 + u)5. [2] (ii) By replacing u with 2x – 5x2, find the coefficient of x2 in the expansion of (2 + 2x – 5x2)5. [4]

4 0606/01/M/J/07 © UCLES 2007 5 A curve has the equation y x x = + 9 . (i) Find expressions for dy dx and d2y dx2 . [4] (ii) Show that the curve has a stationary value when x = 9. [1] (iii) Find the nature of this stationary value. [2] 6 A B O j i The diagram shows a large rectangular television screen in which one corner is taken as the origin O and i and j are unit vectors along two of the edges. In a game, an alien spacecraft appears at the point A with position vector 12j cm and moves across the screen with velocity (40i + 15j) cm per second. A player fires a missile from a point B; the missile is fired 0.5 seconds after the spacecraft appears on the screen. The point B has position vector 46i cm and the velocity of the missile is (ki +30j) cm per second, where k is a constant. Given that the missile hits the spacecraft, (i) show that the spacecraft moved across the screen for 1.8 seconds before impact, [4] (ii) find the value of k. [3] 7 (a) Use the substitution u = 5x to solve the equation 5x + 1 = 8 + 4 (5–x). [5] (b) Given that log(p – q) = log p – log q , express p in terms of q. [3] 8 (a) Solve, for 0  x  2, the equation 1 + 5cos 3x = 0, giving your answer in radians correct to 2 decimal places. [3] (b) Find all the angles between 0º and 360º such that sec y + 5tan y = 3cos y. [5]

5 0606/01/M/J/07 © UCLES 2007 [Turn over 9 x 0.100 0.125 0.160 0.200 0.400 y 0.050 0.064 0.085 0.111 0.286 The table above shows experimental values of the variables x and y. (i) On graph paper draw the graph of 1 y against 1 x . [3] Hence, (ii) express y in terms of x, [4] (iii) find the value of x for which y = 0.15. [2] 10 A B E D C 8m 5m 5m The diagram shows an isosceles triangle ABC in which AB = 8 m, BC = CA = 5 m. ABDA is a sector of the circle, centre A and radius 8 m. CBEC is a sector of the circle, centre C and radius 5 m. (i) Show that angle BCE is 1.287 radians correct to 3 decimal places. [2] (ii) Find the perimeter of the shaded region. [4] (iii) Find the area of the shaded region. [4] [Question 11 is printed on the next page.]

6 0606/01/M/J/07 © UCLES 2007 11 Answer only one of the following two alternatives. EITHER y y = 3 sin x + 4 cos x x O π 2 The graph shows part of the curve y = 3sin x + 4 cos x for 0  x  π 2 radians. (i) Find the coordinates of the maximum point of the curve. [5] (ii) Find the area of the shaded region. [5] OR y y = (3x + 2)2 12 x O B A C The diagram, which is not drawn to scale, shows part of the curve y = 12 (3x + 2)2 , intersecting the y-axis at A. The tangent to the curve at A meets the x-axis at B. The point C lies on the curve and BC is parallel to the y-axis. (i) Find the x-coordinate of B. [4] (ii) Find the area of the shaded region. [6]

8 0606/01/M/J/07 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2007 question paper 0606 ADDITIONAL MATHEMATICS 0606/01 Paper 1, maximum raw mark 80 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2007 0606 01 © UCLES 2007 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Accuracy mark for a correct result or statement independent of method marks. • When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2, 1, 0 means that the candidate can earn anything from 0 to 2.

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2007 0606 01 © UCLES 2007 The following abbreviations may be used in a mark scheme or used on the scripts: AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) Penalties MR -1 A penalty of MR -1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. OW -1,2 This is deducted from A or B marks when essential working is omitted. PA -1 This is deducted from A or B marks in the case of premature approximation. S -1 Occasionally used for persistent slackness – usually discussed at a meeting. EX -1 Applied to A or B marks when extra solutions are offered to a particular equation. Again, this is usually discussed at the meeting.

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2007 0606 01 © UCLES 2007 1. (i) (ii) A ∩ B’ ∩ C (iii) (X∪Y)’ X’∪Y’ B1 [1] B1 [1] B1 B1 [2] co co co co. 2. 2 4 2 − + = x x y dy/dx = 2) 2 ( ) 4 2 ( 2 ) 2 ( − + − − x x x If x = 4, dy/dx = −2 Perpendicular has m = ½ If x = 4, y = 6 → Eqn y − 6 = ½(x − 4) [2y=x+8] M1 A1 M1 B1 A1 [5] Formula must be completely correct co. (may be implied) Independent of first M mark. Anywhere in the question. 3. 18 2 3 + = y x 0 50 2 23 2 2 = + + − y x x → 0 16 10 2 = + − x x or 0 18 3 2 = − + y y M1A1 Complete elimination of x/y for M. → → (2, −6) and (8, 3) DM1 A1 Correct method of solution of quad. Vector moves or other → P (4, −3) M1A1√ [6] Any valid method. 4. (i) 5) 2 ( u + = 32 + 80u + 80u2 (ii) Replaces u by 2x − 5x2 –400 from ‘u’ term or +320 from ‘u2’ term Also … +80 2 2 ) 5 2 ( x x − → − 400 + 320 = −80 B2,1,0 [2] M1 B1 M1 A1√ [4] One lost for each error Recognises and uses the link. Co (may be implied by answer) Needs to look at 2 terms for x² From his original expansion.

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2007 0606 01 © UCLES 2007 5. x x y 9 + = (i) x y d d = 2 3 2 9 2 1 x x − 2 2 dx d y = 2 3 4 1 x − + 2 5 4 27 x (ii) If x = 9 , dx dy = 0 . B1 B1 B1 B1 [4] B1 [1] Accept all these B marks if given as negative powers of x Answer given. (iii) If x = 9, 2 2 dx d y > 0. Minimum M1 A1 [2] Looks at sign of 2 2 dx d y . Needs all correct for the A mark. 6. (i) In 1.8s , alien goes 27 cm up. In 1.3 s missile goes 39 up. But alien starts at 12 up. → 39 – 27 = 12 (ii) In 1.8s. alien goes 72 across In 1.3 s, missile goes 1.3k 72 = 1.3k + 46 → k = 20. B1 B1 M1 A1 [4] B1 M1 A1 [3] Equates 2 vertical displacements. Equates 2 horizontal displacements. 7. (a) ) 5 ( 4 8 5 1 x x − + + = → 1 4 8 5 − + = u u → 5u2 − 8u − 4 = 0 → u = 2 or − 0.4 Soln of 5x = 2 → x = lg2 ÷ lg5 → x = 0.431 (b) q p q p log log ) log( − = − B1 B1 M1 M1 A1 [5] B1 for 5u and B1 for 4u–1 Solution of a quadratic. Allow for any soln of 5x = k. co. = log (p/q) B1 co. p − q = p/q M1 Eliminating lg + good algebra. → 1 2 − = q q p A1 co. [3] 8. (a) 0 3 cos 5 1 = + x cos3x = −0.2 3x = cos–1(−0.2) → x = 0.59 or 1.50 (b) . cos 3 tan 5 sec y y y = + secy = 1/cosy and × cosy uses cos2 = 1 − sin2 → 3sin2y + 5siny − 2 = 0 + solution → sin y = ⅓ y = 19.5º and 160.5º. M1 A1 A1 [3] M1 M1 DM1 A1 A1√ [5] Looks up cos before ÷ 3 co.co. Needs both of these. Needs correct link. Solution of quadratic co. √ for 180 − (first ans)

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2007 0606 01 © UCLES 2007 9. (i) 1/x 10 8 6.25 5 2.5 1/y 20 15.6 11.8 9.0 3.5 (ii) Gradient 2.2 (±0.05) Intercept = −2(±0.1) 2 1 .2.2 1 − = x y → x x y 2 2.2 − = (iii) y = 0.15 1/y = 6.7 → 1/x = 4 → x = 0.254 (±0.010) M1 A2,1,0 [3] B1 B1 M1 A1√ [4] M1 A1 [2] Knows what to do. Accuracy. Within given range – graph needed Uses Y = mX + c Correct form with his m and c. Uses 1/y and 1/x correctly – or solves equation from part (ii). co within range. 10 (i) AC = cos–14/5 = 0.6435 rads BCE = 2×BAC = 1.287 (ii) arc BD = 8×0.6435 =(5.148) arc BE = 5×1.287 = (6.435) DE = 10−8 → Perimeter = sum of these = 13.6 m. M1 A1 [2] M1 B1 DM1 A1 [4] Complete method inc radian use. co – answer given. Any use of s=rθ Anywhere Sum of three parts. co. (iii) Area of ∆ ABC = 3×4 or ½absinC=12 M1 Correct method for triangle. Area of sect CBE = ½×25×1.287= (16.09) M1 Any use of A=½r2θ Area of sect ABD = ½×64×0.6435=(20.59) → shaded area = 12+16.09−20.59 M1 Must be linked correctly. Not DM. → 7.50 m2 A1 Correct to 3 sf. [4]

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2007 0606 01 © UCLES 2007 Q 9

Page 8 Mark Scheme Syllabus Paper IGCSE – May/June 2007 0606 01 © UCLES 2007 11 EITHER (i) dy/dx = 3cosx − 4sinx = 0 when 3cosx − 4sinx=0 tanx = ¾ → x = 0.644 → y = 5.00 (ii) A = ∫ + 2 0 . cos 4 sin 3 π dx x x M1 A1 DM1 DM1 A1 [5] M1 Attempt at differentiation. co. Sets differential to 0. Arrives at tanθ = k. Both x and y needed. Any attempt to integrate = [–3cosx + 4sinx] A1 A1 Each term. = [0+4}–[–3+0] DM1 Correct use of limits – DM0 if “0” left → 7 A1 co [5] 11 OR 2 ) 2 3 ( 12 + = x y (i) dy/dx = −24 × (3x+2)–3 × 3 When x = 0 , dy/dx = −9 At A, x = 0 and y = 3 → B: x = ⅓ (ii) A = ∫ + 3 / 1 0 2 . ) 2 3 ( 12 dx x = [ −12(3x+2)−1 ÷ 3 ] = −4/3 − −2 = ⅔ Area of triangle = ½×3×⅓ = ½ → A = 1/6 B1 B1 B1 B1 [4] M1 A1 A1 DM1 M1 A1 [6] For −24 × (3x+2)–3, for × 3 co. co Attempt needed to integrate For −12(3x+2)−1). For ÷ 3. Not given if bottom limit ignored. Anywhere. co DM1 for quadratic equation. Equation must be set to 0 if using formula or factors. Formula. Factors Must be correct Must attempt to put quadratic into 2 factors. – ignore arithmetic and algebraic slips. Each factor then equated to 0.